Wine Review Analysis

Introduction

In this project we use a dataset of wine reviews to predict review points from numerical, categorical and textual predictors.

The data is from Kaggle Datasets, and covers 150k wine reviews along with some attributes of the wines. It can be found here. (A (free) Kaggle login is required to access it directly from kaggle.com). The data was originally scraped from WineEnthusiast.

The dataset contains the following columns:

• Points: the number of points the wine received in its review, on a scale of 1-100. However, only wines with >= 80 points were published in the dataset. We plan to transform this feature and check out logit, probit, BoxCox and log transforms. This will be our response variable.

• Description: a description of the wine’s taste, smell, look, feel, etc. We plan to use LDA topic modeling to convert the text description into a small vector of numerical predictors. We plan to generate a “sentiment” numeric predictor using a sentiment classifier. This is an additive predictor.

• Price: the cost for a bottle of the wine. We will transform this using log or inverse or a power of the inverse (TBD by CV), or any combination of these.

• Variety: the type of grapes used

• Country: the country that the wine is from

This is a particularly interesting problem for several reasons:

• We need to transform both predictors and response, which makes for a more interesting problem. We can dig into research of optimal transforms using cross validation over a grid of models.
• The regression via glm shows heavy tails on the errors in the QQ plot. Because we have 137k data points, this offers us an opportunity to argue that by the CLT the heavy tails do not obviate our use of t and F tests. We can explore the pitfalls of heavy tails and try to use the “heavy” package to do a glm with heavy tails. Again, this is more interesting than a more straight-forward problem.
• Chance to use LDA topic modeling and sentiment analysis

Methods

Feature Engineering

• The country was used to compute the continentTopFive, which was added as a categorical feature. This feature is the continent of the country of origin of the wine, with the separation of the top 5 wine producers in Europe into their own class. These include France, Germany, Italy, Spain and Portugal. The levels are indicated in the plot of points vs log(price), by continentTopFive.

### Add 'continentTopFive' feature
wine = data.frame(wine, continentTopFive = getContinent_withTop5(wine$country))
#levels(wine$continentTopFive)
ggplot(wine, aes(x=log(price), y=points)) + 
  geom_point(aes(col = continentTopFive)) +stat_summary(fun.data=mean_cl_normal) + 
    stat_smooth(aes(colour=continentTopFive),method="lm",se = FALSE) + 
            scale_colour_manual(values = c("red","green", "blue", "orange", "dodger blue", "violet", "dark green")) + ggtitle("Points versus Log(price), by ContinentTopFive") + theme(plot.title = element_text(hjust = 0.5)) + labs(x="log(price)",y="points")

• LDA for 2 topics was performed on the description of the wine. The probability of being in the first topic was added as a feature named “topic1”.
• The sentiment of the description of the wine was computed using bing. Sentiment was added as a numeric predictor, named “sentiment”, ranging from -6 to 11.

### Perform LDA and sentiment analysis on the description

corpus = tm::Corpus(tm::VectorSource(as.character(wine$description)))
## transform corpus: remove white space and punctuation, transform to lower case, remove stop words, stem.
corpus = tm::tm_map(corpus, stripWhitespace)
corpus = tm::tm_map(corpus, removePunctuation)
corpus = tm::tm_map(corpus, content_transformer(tolower))
corpus = tm::tm_map(corpus, removeWords, stopwords("english"))
# corpus = tm_map(corpus, removeWords, c("wine"))
corpus = tm::tm_map(corpus, stemDocument)
## create DocumentTerm matrix for LDA
corpus_dtm = DocumentTermMatrix(corpus)

NUM_TOPICS = 2

corpus_lda = topicmodels::LDA(corpus_dtm, k = NUM_TOPICS, control = list(seed = 1234))
corpus_documents = tidy(corpus_lda, matrix = "gamma")

SENTIMENT_ENGINE = "bing"
## Derive sentiment score
corpus_sentiments <- tidy(corpus_dtm) %>%
  inner_join(get_sentiments(SENTIMENT_ENGINE), by = c(term = "word")) %>%
  count(document, sentiment, wt = count) %>%
  ungroup() %>%
  spread(sentiment, n, fill = 0) %>%
  mutate(sentiment = positive - negative) %>%
  arrange(sentiment)

## introduce new predictors
wine = data.frame(wine, topic1 = corpus_documents$gamma[1:nrow(wine)], document = rownames(wine))
wine = merge(wine, corpus_sentiments[,c(1,4)], by = "document")
wine = wine[complete.cases(wine[, c("points","price", "continentTopFive", "topic1", "sentiment")]), c("points","price", "continentTopFive", "topic1", "sentiment")]

• The resulting predictors after feature engineering are shown below, with ‘points’ as the response.
• Because price ranges from 4 to 2300 USD, and is positive, ranging over 3 orders of magnitude, log(price) is used as a predictor, instead of price.

names(wine)

## [1] "points"           "price"            "continentTopFive"
## [4] "topic1"           "sentiment"

Training and Test Set Generation

We shuffled our data, then created an 80% training, 20% test split for later use in cross validation. We did 10 fold cross validation of every model.

## REMOVE TEST SET = 20% of data.
#Randomly shuffle the data
wine = wine[sample(nrow(wine)),]

# Sample 80% to training, 20% to test set.
smp_size = floor(0.80 * nrow(wine))
train_ind = sample(seq_len(nrow(wine)), size = smp_size)
wine_train = wine[train_ind, ]
wine_test = wine[-train_ind, ]
# unit test the partitioning
nrow(wine) == nrow(wine_test) + nrow(wine_train)

Model Selection

Model #1

• The model started with the base of: \(log(points-79.999) \sim log(price)*continentTopFive*sentiment*topic1 + I(log(price)^2) + I(log(price)^3)\)
• The optimal model after backward BIC selection was: \(log(points - 79.999) \sim log(price) + continentTopFive + topic1 + log(price):continentTopFive + continentTopFive:topic1\)

mod = lm(log(points-79.999) ~ log(price)*continentTopFive*sentiment*topic1 + I(log(price)^2) + I(log(price)^3), data=wine_train)
select = step(mod, direction="backward", k=log(nrow(wine_train)), trace=FALSE)
# plot(select)
# bptest(select)
# shapiro.test(sample(resid(select),5000))

# remove outliers and highly influential points and refit the *select*ed model
keep = which(abs(resid(select)) <= 7 & cooks.distance(select) <= 4 / nrow(wine_train))

select_formula1 = log(points - 79.999) ~ log(price) + continentTopFive + topic1 + 
    log(price):continentTopFive + continentTopFive:topic1
mod1 = lm(select_formula1, data=wine_train, subset=keep)

# Plot details and perform tests for homoscedasticity and normality of errors.
plot(mod1)

bptest(mod1)

## 
##  studentized Breusch-Pagan test
## 
## data:  mod1
## BP = 2814.5, df = 20, p-value < 2.2e-16

shapiro.test(sample(resid(mod1),5000))

## 
##  Shapiro-Wilk normality test
## 
## data:  sample(resid(mod1), 5000)
## W = 0.91539, p-value < 2.2e-16

wine_train_cv = wine_train[keep,]

RMSE1 = cv_mean_rmse(select_formula1, wine_train_cv, 10, function(x) {exp(x) + 79.999} )

• Outliers, in which the residual had magnitude over 7 points, were removed. Highly influential points were removed and the model was refit. 1266 observations were removed in total.
• 10-fold cross-validation yielded a mean test RMSE of 2.5673186.

Model 2

• The model started with the base of: \(points \sim log(price)*continentTopFive*sentiment*topic1 + I(log(price)^2) + I(log(price)^3)\)
• The optimal model after backward BIC selection was: \(points \sim log(price) + continentTopFive + topic1 + log(price):continentTopFive + log(price):topic1 + continentTopFive:topic1\)

mod = lm(points ~ log(price)*continentTopFive*sentiment*topic1 + I(log(price)^2) + I(log(price)^3), data=wine_train)
select = step(mod, direction="backward", k=log(nrow(wine_train)), trace=FALSE)
# plot(select)
# bptest(select)
# shapiro.test(sample(resid(select),5000))

#remove highly influential points
keep = which(cooks.distance(select) <= 4 / nrow(wine_train))
select_formula2 = points ~ log(price) + continentTopFive + topic1 + log(price):continentTopFive + 
    log(price):topic1 + continentTopFive:topic1
mod2 = lm(select_formula2, data=wine_train, subset=keep)
plot(mod2)

bptest(mod2)

## 
##  studentized Breusch-Pagan test
## 
## data:  mod2
## BP = 5516.9, df = 21, p-value < 2.2e-16

shapiro.test(sample(resid(mod2),5000))

## 
##  Shapiro-Wilk normality test
## 
## data:  sample(resid(mod2), 5000)
## W = 0.998, p-value = 4.588e-06

wine_train_cv = wine_train[keep,]

RMSE2 = cv_mean_rmse(select_formula2, wine_train_cv, 10)

• Highly influential points were removed and the model was refit. 3387 observations were removed in total.
• 10-fold cross-validation yielded a mean test RMSE of 2.4329049.

Model 3

• The Box-Cox transform of the response was performed to try to counteract the indicated heteroskedasticity.
• The model started with a base of: \(bc\_points \sim log(price)*continentTopFive*sentiment*topic1 + I(log(price)^2) + I(log(price)^3)\)
• The optimal model after backward BIC selection was: \(bc\_points \sim log(price) + continentTopFive + topic1 + log(price):continentTopFive + log(price):topic1 + continentTopFive:topic1\)

bc_mod = caret::BoxCoxTrans(wine_train$points)

wine_train = cbind(wine_train, bc_points = predict(bc_mod, wine_train$points))
mod = lm(bc_points ~ log(price)*continentTopFive*sentiment*topic1 + I(log(price)^2) + I(log(price)^3), data=wine_train)
select = step(mod, direction="backward", k=log(nrow(wine_train)), trace=FALSE)

keep = cooks.distance(mod) <= 4 / nrow(wine_train)
select_formula3 = bc_points ~ log(price) + continentTopFive + topic1 + log(price):continentTopFive + 
    log(price):topic1 + continentTopFive:topic1

mod3 = lm(select_formula3, data=wine_train, subset = keep)
plot(mod3)

bptest(mod3)

## 
##  studentized Breusch-Pagan test
## 
## data:  mod3
## BP = 4528.8, df = 21, p-value < 2.2e-16

shapiro.test(sample(resid(mod3),5000))

## 
##  Shapiro-Wilk normality test
## 
## data:  sample(resid(mod3), 5000)
## W = 0.99359, p-value = 3.441e-14

### Perform 10 fold cross validation

#Randomly shuffle the training set and create 10 folds
wine_train_cv = wine_train[keep,]

# see section 13.1.2 at http://daviddalpiaz.github.io/appliedstats/transformations.html#response-transformation
RMSE3 = cv_mean_rmse(select_formula3, wine_train_cv, 10, function(x) {(x*bc_mod$lambda + 1)^(1/bc_mod$lambda)})

• Highly influential points were removed and the model was refit. 0 observations were removed in total.
• 10-fold cross-validation yielded a mean test RMSE of 2.4495482.

Model 4

• Weighted least squares was performed, with weight being proportional to the inverse of the observed variance of points. The observed variance of residuals peaks at points = 90 and tapers to the ends of the points range. The weight function therefore assumes its minimum at points = 90 and increases at the ends of the range of points.
• The weight function used: \(weights = sqrt(1 / 1 + abs(points -90))\)
• The model started with a base of: \(points \sim log(price)*continentTopFive*sentiment*topic1 + I(log(price)^2) + I(log(price)^3)\)
• The optimal model after backward BIC selection was: \(points \sim log(price) + continentTopFive + topic1 + log(price):continentTopFive + log(price):topic1 + continentTopFive:topic1\)

mod = lm(points ~ log(price)*continentTopFive*sentiment*topic1 + I(log(price)^2) + I(log(price)^3), data=wine_train, subset = keep, weights = sqrt(1 / 1 + abs(points -90)))
select = step(mod, direction="backward", k=log(nrow(wine_train)), trace=FALSE)
keep = cooks.distance(mod) <= 4 / nrow(wine_train)

select_formula4 = points ~ log(price) + continentTopFive + topic1 + log(price):continentTopFive + 
    log(price):topic1 + continentTopFive:topic1
mod4 = lm(select_formula4, data=wine_train, subset = keep, weights = sqrt(1 / 1 + abs(points -90)))

plot(mod4)

bptest(mod4)

## 
##  studentized Breusch-Pagan test
## 
## data:  mod4
## BP = 4449.3, df = 21, p-value < 2.2e-16

shapiro.test(sample(resid(mod4),5000))

## 
##  Shapiro-Wilk normality test
## 
## data:  sample(resid(mod4), 5000)
## W = 0.99787, p-value = 2.059e-06

### Perform 10 fold cross validation

#Randomly shuffle the training set and create 10 folds
wine_train_cv = wine_train[keep,]

RMSE4 = cv_mean_rmse(select_formula4, wine_train_cv, 10)

• Highly influential points were removed and the model was refit. 3125 observations were removed in total.
• 10-fold cross-validation yielded a mean test RMSE of 2.53495.

Model 5

• The model started with the base of: \(points \sim log(price)*continentTopFive*sentiment*topic1 + I(log(price)^2) + I(log(price)^3) + I(log(price)^4) + I(log(price)^5) + I(log(price)^6) + I(log(price)^7)\)
• The optimal model after backward BIC selection was: \(points \sim log(price) + continentTopFive + topic1 + log(price):continentTopFive + log(price):topic1 + continentTopFive:topic1\)

mod = lm(points ~ log(price)*continentTopFive*sentiment*topic1 + I(log(price)^2) + I(log(price)^3)  + I(log(price)^4) + I(log(price)^5)  + I(log(price)^6) + I(log(price)^7), data=wine_train)
select = step(mod, direction="backward", k=log(nrow(wine_train)), trace=FALSE)
# plot(select)
# bptest(select)
# shapiro.test(sample(resid(select),5000))

#remove highly influential points
keep = which(cooks.distance(select) <= 4 / nrow(wine_train))

#update
select_formula5 = points ~ log(price) + continentTopFive + topic1 + log(price):continentTopFive + 
    log(price):topic1 + continentTopFive:topic1
mod5 = lm(select_formula5, data=wine_train, subset=keep)
plot(mod2)

bptest(mod2)

## 
##  studentized Breusch-Pagan test
## 
## data:  mod2
## BP = 5516.9, df = 21, p-value < 2.2e-16

shapiro.test(sample(resid(mod5),5000))

## 
##  Shapiro-Wilk normality test
## 
## data:  sample(resid(mod5), 5000)
## W = 0.9943, p-value = 3.436e-13

wine_train_cv = wine_train[keep,]

RMSE5 = cv_mean_rmse(select_formula5, wine_train_cv, 10)

• Highly influential points were removed and the model was refit. 3120 observations were removed in total.
• 10-fold cross-validation yielded a mean test RMSE of 2.4374888.

Results

• 4 models were analyzed as candidates for optimal prediction of points.
• For each, a base model was constructed and backward BIC selection was used to trim the model.
• As seen by the plots and the BP and Shapiro-Wilk tests, normality of the errors is suspect, and heteroskedasticity is indicated for all models. In particular, the variance seems highest for points around 90, and tapers to the extremes. Note however, that since Shapiro-Wilk in R can take a maximum of 5000 residuals, sampling was required, and this test is then dependent on the choice of random seed used in the sampling.
• Many transforms (the 4 here, and not shown) were attempted in order to overcome these violations of the assumptions of multiple linear regression by least squares minimization.
• 10 fold cross-validation was performed on all models, with mean CV RMSE given as below.

rmse_data = data.frame(rmse = c(RMSE1, RMSE2, RMSE3, RMSE4, RMSE5))
colnames(rmse_data) = c("CV RMSE")
rownames(rmse_data) = c("model1", "model2", "model3", "model4", "model5")

best_model = which.min(rmse_data$`CV RMSE`)
best_cv_rmse = round(rmse_data[best_model, 1],4)
eqn1 = paste("$\\hat{y} \\pm 2*", best_cv_rmse, "$", sep = "")

rmse_data

##         CV RMSE
## model1 2.567319
## model2 2.432905
## model3 2.449548
## model4 2.534950
## model5 2.437489

• Model 2 wins by minimal CV RMSE(2.4329).
• The CV RMSE can be used to construct prediction intervals, in lieu of analytic formulae based on normal and homoskedastic errors, as \(\hat{y} \pm 2*2.4329\) for the prediction interval.

preds = predict(mod2, newdata=wine_test)
success_ratio = sum(wine_test$points >= preds - 2*best_cv_rmse & wine_test$points <= preds + 2*best_cv_rmse) / nrow(wine_test)

• Using Model 2 with this prediction interval resulted in a success ratio of 0.9434145 on the withheld test set (i.e. this percentage of the withheld test set points were within the prediction interval of Model #2).

Discussion

We used four different strategies to attempt to meet the assumptions of multiple linear regression and minimize cross-validated RMSE. Attempts include transforming response with BoxCox and log, integrated polynomial models over the engineered features, and weighted least squares regression. In the end, the non-transformed response model had the lowest cross-validated RMSE with a value of 2.4329049.

Normality of the errors and homoskedasticity are suspect, as demonstrated by the Breusch-Pagan and Shapiro-Wilk tests. we tried many attempts to get normality and homoskedasticity (including additional less interesting attempts not shown here), but were not successful. We see that the variance seems highest for points around 90, and tapers to the extremes. However, because Shapiro-Wilk in R can take only a maximum of 5000 residuals, sampling is required, and this test is then dependent on the choice of random seed used in the sampling.

We get around the normal and homoskedastic assumptions by using cross-validated RMSE to create our prediction intervals. Doing so gives a 94% success rate on the withheld test set with a +/- 4.8 point prediction interval.

Potential for future improvements: This project could potentially be further expanded in the future by using deeper NLP, expanding with additional predictors, and expanded model selection attempts. It may also be the case that the data is noisy, with a high base \(\epsilon\) variance of points no matter which predictors are used. (subjective human judgement after all).

Appendix

About the LDA

• Since we did not want to wander too far into topic modelling of the wine descriptions, an LDA with only 2 topics was performed. Future extension to this work should try to mine topics and sentiment further from the sommelier description, via NLP techniques.
• Our LDA treatment relies on 2 references regarding the ‘topicmodels’ library of R:
  • http://tidytextmining.com/topicmodeling.html
  • https://cran.r-project.org/web/packages/topicmodels/vignettes/topicmodels.pdf
• LDA with 2 topics creates 2 probability distributions of words discovered in the processed wine descriptions (after removing punctuation, white space, casting to lower case and stemming, which means taking word roots). Each of these probability distributions is a “topic”. Each description receives a probability of coming from topic1, with the probability of coming from topic2 being 1 minus this value. This probability of being generated from topic 1 is our “topic1” predictor. Our “topic1” predictor remained significant as an additive and integrated predictor in all BIC reduced models.
• The “beta” of a word in a topic is its probability within the topic probability distribution. Below, topic 1 and topic 2 probabilities for various words are displayed, as is the log ratio of the topic probabilities for some words.

corpus_topics = tidy(corpus_lda, matrix = "beta")
#corpus_topics
corpus_top_terms = corpus_topics %>%
  group_by(topic) %>%
  top_n(10, beta) %>%
  ungroup() %>%
  arrange(topic, -beta)

corpus_top_terms %>%
  mutate(term = reorder(term, beta)) %>%
  ggplot(aes(term, beta, fill = factor(topic))) + ggtitle("Word Probabilities for the 2 Topics") +
  geom_col(show.legend = FALSE) +
  facet_wrap(~ topic, scales = "free") +
  coord_flip()

beta_spread = corpus_topics %>%
  mutate(topic = paste0("topic", topic)) %>%
  spread(topic, beta) %>%
  filter(topic1 > .001 | topic2 > .001) %>%
  mutate(log_ratio = log2(topic2 / topic1))
#beta_spread

## Plot the beta spread of the topics
ggplot(beta_spread[order(abs(beta_spread$log_ratio)),][1:30,], aes(x=term, y=log_ratio)) + geom_bar(stat="identity", fill="green", width=.2) + coord_flip() + ggtitle("Log ratio of word probabilities in topic2/topic1")